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2 years ago

How much work is done on 10.0 c of charge to move it through a potential diffrence of 9.0 v in 10.0s

Physics

1 answer:

GenaCL600 [577]2 years ago

4 0

We need more evidence to be provided

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How is heat converted into work in a steam engine?

Dmitry_Shevchenko [17]

In an extremely simplified explanation,
What happens is that the steam (heat) that is generated by the burning of coal is used to rotate the motor of the steam engine.

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3 years ago

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string is oscillated at 440 Hz, which is the frequency of t

Katarina [22]

Answer:

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Explanation:

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2 years ago

A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh

MrMuchimi

Answer:

<em>W=700 Joule</em>

Explanation:

<u>Physics Work </u>

Is the dot product of the force vector by the displacement vector

$W=\vec F \cdot \vec r$

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

$W=F.d$

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

$W=350\ N\ .\ 2\ m=700\ Joule$

4 0

3 years ago

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

8 0

2 years ago

I need help, please answer

Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0

3 years ago